Chapter 14, Problem 46E

Question

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Answer 1

Question

Chapter 14, Problem 46E

Expert Solution & Answer

To determine

To find:

• The range of the given projectile equations.

• The graph of the projectile at suitable x values.

• The graph of the height versus time.

• The range of the given projectile equations.

Answer to Problem 46E

Solution:

• The required range is 0.35

• The required graph is stated as follows.

• The required apex is 0.1.

Explanation of Solution

• The range of the given projectile equations.

The given equations are,

x=v0cos(θ0)ty=v0sin(θ0)t−12gt2

The formula to calculate the range is,

range=v02sin(2θ0)g

Here, v0 is the initial velocity, θ0 is the angle of departure, g is the acceleration due to gravity and t is the time.

Let v0 be 2 m/s, θ0 be 30°, t be 1 sec and g be 9.81 m/s.

Substitute 2 m/s for v0, 30° for θ0, 1 sec for t and 9.81 m/s for g in the above equation.

range=v02sin(2θ0)g=22sin(2×30°)9.81=0.35

MATLAB Code:

t = 0:0.001:2;

v0 = 2;

g = 9.81;

theta = 30;

x = v0.*cosd(theta).*t;

y = v0.*sind(theta).*t -0.5.*g.*t.*t;

range = v0.^2.*sind(2*theta)/g

Save the MATLAB function with name chapter14_54793_14_46_1E.m in the current folder. Execute the function by typing the function name at the command window to generate output.

Result:

MATLAB: A Practical Introduction to Programming and Problem Solving, Chapter 14, Problem 46E , additional homework tip 1

Therefore, the result is stated above.

• The graph of the projectile at suitable x values.

MATLAB Code:

t = 0:0.001:2;

v0 = 2;

g = 9.81;

theta = 30;

x = v0.*cosd(theta).*t;

y = v0.*sind(theta).*t -0.5.*g.*t.*t;

plot(t, x)

title('Position(x) vs Time(t)')

Save the MATLAB function with name chapter14_54793_14_46_2E.m in the current folder. Execute the function by typing the function name at the command window to generate output.

Result:

MATLAB: A Practical Introduction to Programming and Problem Solving, Chapter 14, Problem 46E , additional homework tip 2

Therefore, the result is stated above.

• The graph of the height versus time.

MATLAB Code:

t = 0:0.001:2;

v0 = 2;

g = 9.81;

theta = 30;

x = v0.*cosd(theta).*t;

y = v0.*sind(theta).*t -0.5.*g.*t.*t;

plot(t, y)

title('Height(y) vs Time(t)')

Save the MATLAB function with name chapter14_54793_14_46_3E.m in the current folder. Execute the function by typing the function name at the command window to generate output.

Result:

MATLAB: A Practical Introduction to Programming and Problem Solving, Chapter 14, Problem 46E , additional homework tip 3

Therefore, the result is stated above.

• The range of the given projectile equations.

The given equations are,

x=v0cos(θ0)ty=v0sin(θ0)t−12gt2

The formula to calculate the maximum time, tmax is,

tmax=v0sin(θ0)g

Here, v0 is the initial velocity, θ0 is the angle of departure, g is the acceleration due to gravity and t is the time.

Let v0 be 2 m/s, θ0 be 30°, t be 1 sec and g be 9.81 m/s.

Substitute 2 m/s for v0, 30° for θ0, and 9.81 m/s for g in the above equation.

tmax=v0sin(θ0)g=2sin(30°)9.81=0.10 sec

MATLAB Code:

t = 0:0.001:2;

v0 = 2;

g = 9.81;

theta = 30;

x = v0.*cosd(theta).*t;

y = v0.*sind(theta).*t -0.5.*g.*t.*t;

apex_time = v0*sind(theta)/g;

fprintf('The apex is %.1f\n', apex_time)

Save the MATLAB function with name chapter14_54793_14_46_4E.m in the current folder. Execute the function by typing the function name at the command window to generate output.

Result:

MATLAB: A Practical Introduction to Programming and Problem Solving, Chapter 14, Problem 46E , additional homework tip 4

Therefore, the required apex is 0.1.

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Answer 2

Question

Chapter 14, Problem 46E

Expert Solution & Answer

To determine

To find:

• The range of the given projectile equations.

• The graph of the projectile at suitable x values.

• The graph of the height versus time.

• The range of the given projectile equations.

Answer to Problem 46E

Solution:

• The required range is 0.35

• The required graph is stated as follows.

• The required apex is 0.1.

Explanation of Solution

• The range of the given projectile equations.

The given equations are,

x=v0cos(θ0)ty=v0sin(θ0)t−12gt2

The formula to calculate the range is,

range=v02sin(2θ0)g

Here, v0 is the initial velocity, θ0 is the angle of departure, g is the acceleration due to gravity and t is the time.

Let v0 be 2 m/s, θ0 be 30°, t be 1 sec and g be 9.81 m/s.

Substitute 2 m/s for v0, 30° for θ0, 1 sec for t and 9.81 m/s for g in the above equation.

range=v02sin(2θ0)g=22sin(2×30°)9.81=0.35

MATLAB Code:

t = 0:0.001:2;

v0 = 2;

g = 9.81;

theta = 30;

x = v0.*cosd(theta).*t;

y = v0.*sind(theta).*t -0.5.*g.*t.*t;

range = v0.^2.*sind(2*theta)/g

Save the MATLAB function with name chapter14_54793_14_46_1E.m in the current folder. Execute the function by typing the function name at the command window to generate output.

Result:

MATLAB: A Practical Introduction to Programming and Problem Solving, Chapter 14, Problem 46E , additional homework tip 1

Therefore, the result is stated above.

• The graph of the projectile at suitable x values.

MATLAB Code:

t = 0:0.001:2;

v0 = 2;

g = 9.81;

theta = 30;

x = v0.*cosd(theta).*t;

y = v0.*sind(theta).*t -0.5.*g.*t.*t;

plot(t, x)

title('Position(x) vs Time(t)')

Save the MATLAB function with name chapter14_54793_14_46_2E.m in the current folder. Execute the function by typing the function name at the command window to generate output.

Result:

MATLAB: A Practical Introduction to Programming and Problem Solving, Chapter 14, Problem 46E , additional homework tip 2

Therefore, the result is stated above.

• The graph of the height versus time.

MATLAB Code:

t = 0:0.001:2;

v0 = 2;

g = 9.81;

theta = 30;

x = v0.*cosd(theta).*t;

y = v0.*sind(theta).*t -0.5.*g.*t.*t;

plot(t, y)

title('Height(y) vs Time(t)')

Save the MATLAB function with name chapter14_54793_14_46_3E.m in the current folder. Execute the function by typing the function name at the command window to generate output.

Result:

MATLAB: A Practical Introduction to Programming and Problem Solving, Chapter 14, Problem 46E , additional homework tip 3

Therefore, the result is stated above.

• The range of the given projectile equations.

The given equations are,

x=v0cos(θ0)ty=v0sin(θ0)t−12gt2

The formula to calculate the maximum time, tmax is,

tmax=v0sin(θ0)g

Here, v0 is the initial velocity, θ0 is the angle of departure, g is the acceleration due to gravity and t is the time.

Let v0 be 2 m/s, θ0 be 30°, t be 1 sec and g be 9.81 m/s.

Substitute 2 m/s for v0, 30° for θ0, and 9.81 m/s for g in the above equation.

tmax=v0sin(θ0)g=2sin(30°)9.81=0.10 sec

MATLAB Code:

t = 0:0.001:2;

v0 = 2;

g = 9.81;

theta = 30;

x = v0.*cosd(theta).*t;

y = v0.*sind(theta).*t -0.5.*g.*t.*t;

apex_time = v0*sind(theta)/g;

fprintf('The apex is %.1f\n', apex_time)

Save the MATLAB function with name chapter14_54793_14_46_4E.m in the current folder. Execute the function by typing the function name at the command window to generate output.

Result:

MATLAB: A Practical Introduction to Programming and Problem Solving, Chapter 14, Problem 46E , additional homework tip 4

Therefore, the required apex is 0.1.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 14, Problem 46E

Expert Solution & Answer

To determine

To find:

• The range of the given projectile equations.

• The graph of the projectile at suitable x values.

• The graph of the height versus time.

• The range of the given projectile equations.

Answer to Problem 46E

Solution:

• The required range is 0.35

• The required graph is stated as follows.

• The required apex is 0.1.

Explanation of Solution

• The range of the given projectile equations.

The given equations are,

x=v0cos(θ0)ty=v0sin(θ0)t−12gt2

The formula to calculate the range is,

range=v02sin(2θ0)g

Here, v0 is the initial velocity, θ0 is the angle of departure, g is the acceleration due to gravity and t is the time.

Let v0 be 2 m/s, θ0 be 30°, t be 1 sec and g be 9.81 m/s.

Substitute 2 m/s for v0, 30° for θ0, 1 sec for t and 9.81 m/s for g in the above equation.

range=v02sin(2θ0)g=22sin(2×30°)9.81=0.35

MATLAB Code:

t = 0:0.001:2;

v0 = 2;

g = 9.81;

theta = 30;

x = v0.*cosd(theta).*t;

y = v0.*sind(theta).*t -0.5.*g.*t.*t;

range = v0.^2.*sind(2*theta)/g

Save the MATLAB function with name chapter14_54793_14_46_1E.m in the current folder. Execute the function by typing the function name at the command window to generate output.

Result:

MATLAB: A Practical Introduction to Programming and Problem Solving, Chapter 14, Problem 46E , additional homework tip 1

Therefore, the result is stated above.

• The graph of the projectile at suitable x values.

MATLAB Code:

t = 0:0.001:2;

v0 = 2;

g = 9.81;

theta = 30;

x = v0.*cosd(theta).*t;

y = v0.*sind(theta).*t -0.5.*g.*t.*t;

plot(t, x)

title('Position(x) vs Time(t)')

Save the MATLAB function with name chapter14_54793_14_46_2E.m in the current folder. Execute the function by typing the function name at the command window to generate output.

Result:

MATLAB: A Practical Introduction to Programming and Problem Solving, Chapter 14, Problem 46E , additional homework tip 2

Therefore, the result is stated above.

• The graph of the height versus time.

MATLAB Code:

t = 0:0.001:2;

v0 = 2;

g = 9.81;

theta = 30;

x = v0.*cosd(theta).*t;

y = v0.*sind(theta).*t -0.5.*g.*t.*t;

plot(t, y)

title('Height(y) vs Time(t)')

Save the MATLAB function with name chapter14_54793_14_46_3E.m in the current folder. Execute the function by typing the function name at the command window to generate output.

Result:

MATLAB: A Practical Introduction to Programming and Problem Solving, Chapter 14, Problem 46E , additional homework tip 3

Therefore, the result is stated above.

• The range of the given projectile equations.

The given equations are,

x=v0cos(θ0)ty=v0sin(θ0)t−12gt2

The formula to calculate the maximum time, tmax is,

tmax=v0sin(θ0)g

Here, v0 is the initial velocity, θ0 is the angle of departure, g is the acceleration due to gravity and t is the time.

Let v0 be 2 m/s, θ0 be 30°, t be 1 sec and g be 9.81 m/s.

Substitute 2 m/s for v0, 30° for θ0, and 9.81 m/s for g in the above equation.

tmax=v0sin(θ0)g=2sin(30°)9.81=0.10 sec

MATLAB Code:

t = 0:0.001:2;

v0 = 2;

g = 9.81;

theta = 30;

x = v0.*cosd(theta).*t;

y = v0.*sind(theta).*t -0.5.*g.*t.*t;

apex_time = v0*sind(theta)/g;

fprintf('The apex is %.1f\n', apex_time)

Save the MATLAB function with name chapter14_54793_14_46_4E.m in the current folder. Execute the function by typing the function name at the command window to generate output.

Result:

MATLAB: A Practical Introduction to Programming and Problem Solving, Chapter 14, Problem 46E , additional homework tip 4

Therefore, the required apex is 0.1.

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Answer 4

Question

Chapter 14, Problem 46E

Expert Solution & Answer

To determine

To find:

• The range of the given projectile equations.

• The graph of the projectile at suitable x values.

• The graph of the height versus time.

• The range of the given projectile equations.

Answer to Problem 46E

Solution:

• The required range is 0.35

• The required graph is stated as follows.

• The required apex is 0.1.

Explanation of Solution

• The range of the given projectile equations.

The given equations are,

x=v0cos(θ0)ty=v0sin(θ0)t−12gt2

The formula to calculate the range is,

range=v02sin(2θ0)g

Here, v0 is the initial velocity, θ0 is the angle of departure, g is the acceleration due to gravity and t is the time.

Let v0 be 2 m/s, θ0 be 30°, t be 1 sec and g be 9.81 m/s.

Substitute 2 m/s for v0, 30° for θ0, 1 sec for t and 9.81 m/s for g in the above equation.

range=v02sin(2θ0)g=22sin(2×30°)9.81=0.35

MATLAB Code:

t = 0:0.001:2;

v0 = 2;

g = 9.81;

theta = 30;

x = v0.*cosd(theta).*t;

y = v0.*sind(theta).*t -0.5.*g.*t.*t;

range = v0.^2.*sind(2*theta)/g

Save the MATLAB function with name chapter14_54793_14_46_1E.m in the current folder. Execute the function by typing the function name at the command window to generate output.

Result:

MATLAB: A Practical Introduction to Programming and Problem Solving, Chapter 14, Problem 46E , additional homework tip 1

Therefore, the result is stated above.

• The graph of the projectile at suitable x values.

MATLAB Code:

t = 0:0.001:2;

v0 = 2;

g = 9.81;

theta = 30;

x = v0.*cosd(theta).*t;

y = v0.*sind(theta).*t -0.5.*g.*t.*t;

plot(t, x)

title('Position(x) vs Time(t)')

Save the MATLAB function with name chapter14_54793_14_46_2E.m in the current folder. Execute the function by typing the function name at the command window to generate output.

Result:

MATLAB: A Practical Introduction to Programming and Problem Solving, Chapter 14, Problem 46E , additional homework tip 2

Therefore, the result is stated above.

• The graph of the height versus time.

MATLAB Code:

t = 0:0.001:2;

v0 = 2;

g = 9.81;

theta = 30;

x = v0.*cosd(theta).*t;

y = v0.*sind(theta).*t -0.5.*g.*t.*t;

plot(t, y)

title('Height(y) vs Time(t)')

Save the MATLAB function with name chapter14_54793_14_46_3E.m in the current folder. Execute the function by typing the function name at the command window to generate output.

Result:

MATLAB: A Practical Introduction to Programming and Problem Solving, Chapter 14, Problem 46E , additional homework tip 3

Therefore, the result is stated above.

• The range of the given projectile equations.

The given equations are,

x=v0cos(θ0)ty=v0sin(θ0)t−12gt2

The formula to calculate the maximum time, tmax is,

tmax=v0sin(θ0)g

Here, v0 is the initial velocity, θ0 is the angle of departure, g is the acceleration due to gravity and t is the time.

Let v0 be 2 m/s, θ0 be 30°, t be 1 sec and g be 9.81 m/s.

Substitute 2 m/s for v0, 30° for θ0, and 9.81 m/s for g in the above equation.

tmax=v0sin(θ0)g=2sin(30°)9.81=0.10 sec

MATLAB Code:

t = 0:0.001:2;

v0 = 2;

g = 9.81;

theta = 30;

x = v0.*cosd(theta).*t;

y = v0.*sind(theta).*t -0.5.*g.*t.*t;

apex_time = v0*sind(theta)/g;

fprintf('The apex is %.1f\n', apex_time)

Save the MATLAB function with name chapter14_54793_14_46_4E.m in the current folder. Execute the function by typing the function name at the command window to generate output.

Result:

MATLAB: A Practical Introduction to Programming and Problem Solving, Chapter 14, Problem 46E , additional homework tip 4

Therefore, the required apex is 0.1.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Expert Solution & Answer

To determine

To find:

• The range of the given projectile equations.

• The graph of the projectile at suitable x values.

• The graph of the height versus time.

• The range of the given projectile equations.

Answer to Problem 46E

Solution:

• The required range is 0.35

• The required graph is stated as follows.

• The required apex is 0.1.

Explanation of Solution

• The range of the given projectile equations.

The given equations are,

x=v0cos(θ0)ty=v0sin(θ0)t−12gt2

The formula to calculate the range is,

range=v02sin(2θ0)g

Here, v0 is the initial velocity, θ0 is the angle of departure, g is the acceleration due to gravity and t is the time.

Let v0 be 2 m/s, θ0 be 30°, t be 1 sec and g be 9.81 m/s.

Substitute 2 m/s for v0, 30° for θ0, 1 sec for t and 9.81 m/s for g in the above equation.

range=v02sin(2θ0)g=22sin(2×30°)9.81=0.35

MATLAB Code:

t = 0:0.001:2;

v0 = 2;

g = 9.81;

theta = 30;

x = v0.*cosd(theta).*t;

y = v0.*sind(theta).*t -0.5.*g.*t.*t;

range = v0.^2.*sind(2*theta)/g

Save the MATLAB function with name chapter14_54793_14_46_1E.m in the current folder. Execute the function by typing the function name at the command window to generate output.

Result:

MATLAB: A Practical Introduction to Programming and Problem Solving, Chapter 14, Problem 46E , additional homework tip 1

Therefore, the result is stated above.

• The graph of the projectile at suitable x values.

MATLAB Code:

t = 0:0.001:2;

v0 = 2;

g = 9.81;

theta = 30;

x = v0.*cosd(theta).*t;

y = v0.*sind(theta).*t -0.5.*g.*t.*t;

plot(t, x)

title('Position(x) vs Time(t)')

Save the MATLAB function with name chapter14_54793_14_46_2E.m in the current folder. Execute the function by typing the function name at the command window to generate output.

Result:

MATLAB: A Practical Introduction to Programming and Problem Solving, Chapter 14, Problem 46E , additional homework tip 2

Therefore, the result is stated above.

• The graph of the height versus time.

MATLAB Code:

t = 0:0.001:2;

v0 = 2;

g = 9.81;

theta = 30;

x = v0.*cosd(theta).*t;

y = v0.*sind(theta).*t -0.5.*g.*t.*t;

plot(t, y)

title('Height(y) vs Time(t)')

Save the MATLAB function with name chapter14_54793_14_46_3E.m in the current folder. Execute the function by typing the function name at the command window to generate output.

Result:

MATLAB: A Practical Introduction to Programming and Problem Solving, Chapter 14, Problem 46E , additional homework tip 3

Therefore, the result is stated above.

• The range of the given projectile equations.

The given equations are,

x=v0cos(θ0)ty=v0sin(θ0)t−12gt2

The formula to calculate the maximum time, tmax is,

tmax=v0sin(θ0)g

Here, v0 is the initial velocity, θ0 is the angle of departure, g is the acceleration due to gravity and t is the time.

Let v0 be 2 m/s, θ0 be 30°, t be 1 sec and g be 9.81 m/s.

Substitute 2 m/s for v0, 30° for θ0, and 9.81 m/s for g in the above equation.

tmax=v0sin(θ0)g=2sin(30°)9.81=0.10 sec

MATLAB Code:

t = 0:0.001:2;

v0 = 2;

g = 9.81;

theta = 30;

x = v0.*cosd(theta).*t;

y = v0.*sind(theta).*t -0.5.*g.*t.*t;

apex_time = v0*sind(theta)/g;

fprintf('The apex is %.1f\n', apex_time)

Save the MATLAB function with name chapter14_54793_14_46_4E.m in the current folder. Execute the function by typing the function name at the command window to generate output.

Result:

MATLAB: A Practical Introduction to Programming and Problem Solving, Chapter 14, Problem 46E , additional homework tip 4

Therefore, the required apex is 0.1.

Answer 6

Expert Solution & Answer

To determine

To find:

• The range of the given projectile equations.

• The graph of the projectile at suitable x values.

• The graph of the height versus time.

• The range of the given projectile equations.

Answer to Problem 46E

Solution:

• The required range is 0.35

• The required graph is stated as follows.

• The required apex is 0.1.

Explanation of Solution

• The range of the given projectile equations.

The given equations are,

x=v0cos(θ0)ty=v0sin(θ0)t−12gt2

The formula to calculate the range is,

range=v02sin(2θ0)g

Here, v0 is the initial velocity, θ0 is the angle of departure, g is the acceleration due to gravity and t is the time.

Let v0 be 2 m/s, θ0 be 30°, t be 1 sec and g be 9.81 m/s.

Substitute 2 m/s for v0, 30° for θ0, 1 sec for t and 9.81 m/s for g in the above equation.

range=v02sin(2θ0)g=22sin(2×30°)9.81=0.35

MATLAB Code:

t = 0:0.001:2;

v0 = 2;

g = 9.81;

theta = 30;

x = v0.*cosd(theta).*t;

y = v0.*sind(theta).*t -0.5.*g.*t.*t;

range = v0.^2.*sind(2*theta)/g

Save the MATLAB function with name chapter14_54793_14_46_1E.m in the current folder. Execute the function by typing the function name at the command window to generate output.

Result:

MATLAB: A Practical Introduction to Programming and Problem Solving, Chapter 14, Problem 46E , additional homework tip 1

Therefore, the result is stated above.

• The graph of the projectile at suitable x values.

MATLAB Code:

t = 0:0.001:2;

v0 = 2;

g = 9.81;

theta = 30;

x = v0.*cosd(theta).*t;

y = v0.*sind(theta).*t -0.5.*g.*t.*t;

plot(t, x)

title('Position(x) vs Time(t)')

Save the MATLAB function with name chapter14_54793_14_46_2E.m in the current folder. Execute the function by typing the function name at the command window to generate output.

Result:

MATLAB: A Practical Introduction to Programming and Problem Solving, Chapter 14, Problem 46E , additional homework tip 2

Therefore, the result is stated above.

• The graph of the height versus time.

MATLAB Code:

t = 0:0.001:2;

v0 = 2;

g = 9.81;

theta = 30;

x = v0.*cosd(theta).*t;

y = v0.*sind(theta).*t -0.5.*g.*t.*t;

plot(t, y)

title('Height(y) vs Time(t)')

Save the MATLAB function with name chapter14_54793_14_46_3E.m in the current folder. Execute the function by typing the function name at the command window to generate output.

Result:

MATLAB: A Practical Introduction to Programming and Problem Solving, Chapter 14, Problem 46E , additional homework tip 3

Therefore, the result is stated above.

• The range of the given projectile equations.

The given equations are,

x=v0cos(θ0)ty=v0sin(θ0)t−12gt2

The formula to calculate the maximum time, tmax is,

tmax=v0sin(θ0)g

Here, v0 is the initial velocity, θ0 is the angle of departure, g is the acceleration due to gravity and t is the time.

Let v0 be 2 m/s, θ0 be 30°, t be 1 sec and g be 9.81 m/s.

Substitute 2 m/s for v0, 30° for θ0, and 9.81 m/s for g in the above equation.

tmax=v0sin(θ0)g=2sin(30°)9.81=0.10 sec

MATLAB Code:

t = 0:0.001:2;

v0 = 2;

g = 9.81;

theta = 30;

x = v0.*cosd(theta).*t;

y = v0.*sind(theta).*t -0.5.*g.*t.*t;

apex_time = v0*sind(theta)/g;

fprintf('The apex is %.1f\n', apex_time)

Save the MATLAB function with name chapter14_54793_14_46_4E.m in the current folder. Execute the function by typing the function name at the command window to generate output.

Result:

MATLAB: A Practical Introduction to Programming and Problem Solving, Chapter 14, Problem 46E , additional homework tip 4

Therefore, the required apex is 0.1.

Answer 7

Expert Solution & Answer

Answer 8

Expert Solution & Answer

Answer 9

To determine

To find:

• The range of the given projectile equations.

• The graph of the projectile at suitable x values.

• The graph of the height versus time.

• The range of the given projectile equations.

Answer 10

Answer to Problem 46E

Solution:

• The required range is 0.35

• The required graph is stated as follows.

• The required apex is 0.1.

Answer 11

Explanation of Solution

• The range of the given projectile equations.

The given equations are,

x=v0cos(θ0)ty=v0sin(θ0)t−12gt2

The formula to calculate the range is,

range=v02sin(2θ0)g

Here, v0 is the initial velocity, θ0 is the angle of departure, g is the acceleration due to gravity and t is the time.

Let v0 be 2 m/s, θ0 be 30°, t be 1 sec and g be 9.81 m/s.

Substitute 2 m/s for v0, 30° for θ0, 1 sec for t and 9.81 m/s for g in the above equation.

range=v02sin(2θ0)g=22sin(2×30°)9.81=0.35

MATLAB Code:

t = 0:0.001:2;

v0 = 2;

g = 9.81;

theta = 30;

x = v0.*cosd(theta).*t;

y = v0.*sind(theta).*t -0.5.*g.*t.*t;

range = v0.^2.*sind(2*theta)/g

Save the MATLAB function with name chapter14_54793_14_46_1E.m in the current folder. Execute the function by typing the function name at the command window to generate output.

Result:

MATLAB: A Practical Introduction to Programming and Problem Solving, Chapter 14, Problem 46E , additional homework tip 1

Therefore, the result is stated above.

• The graph of the projectile at suitable x values.

MATLAB Code:

t = 0:0.001:2;

v0 = 2;

g = 9.81;

theta = 30;

x = v0.*cosd(theta).*t;

y = v0.*sind(theta).*t -0.5.*g.*t.*t;

plot(t, x)

title('Position(x) vs Time(t)')

Save the MATLAB function with name chapter14_54793_14_46_2E.m in the current folder. Execute the function by typing the function name at the command window to generate output.

Result:

MATLAB: A Practical Introduction to Programming and Problem Solving, Chapter 14, Problem 46E , additional homework tip 2

Therefore, the result is stated above.

• The graph of the height versus time.

MATLAB Code:

t = 0:0.001:2;

v0 = 2;

g = 9.81;

theta = 30;

x = v0.*cosd(theta).*t;

y = v0.*sind(theta).*t -0.5.*g.*t.*t;

plot(t, y)

title('Height(y) vs Time(t)')

Save the MATLAB function with name chapter14_54793_14_46_3E.m in the current folder. Execute the function by typing the function name at the command window to generate output.

Result:

MATLAB: A Practical Introduction to Programming and Problem Solving, Chapter 14, Problem 46E , additional homework tip 3

Therefore, the result is stated above.

• The range of the given projectile equations.

The given equations are,

x=v0cos(θ0)ty=v0sin(θ0)t−12gt2

The formula to calculate the maximum time, tmax is,

tmax=v0sin(θ0)g

Here, v0 is the initial velocity, θ0 is the angle of departure, g is the acceleration due to gravity and t is the time.

Let v0 be 2 m/s, θ0 be 30°, t be 1 sec and g be 9.81 m/s.

Substitute 2 m/s for v0, 30° for θ0, and 9.81 m/s for g in the above equation.

tmax=v0sin(θ0)g=2sin(30°)9.81=0.10 sec

MATLAB Code:

t = 0:0.001:2;

v0 = 2;

g = 9.81;

theta = 30;

x = v0.*cosd(theta).*t;

y = v0.*sind(theta).*t -0.5.*g.*t.*t;

apex_time = v0*sind(theta)/g;

fprintf('The apex is %.1f\n', apex_time)

Save the MATLAB function with name chapter14_54793_14_46_4E.m in the current folder. Execute the function by typing the function name at the command window to generate output.

Result:

MATLAB: A Practical Introduction to Programming and Problem Solving, Chapter 14, Problem 46E , additional homework tip 4

Therefore, the required apex is 0.1.